Problem: The grades on a geometry midterm at Springer are normally distributed with $\mu = 73$ and $\sigma = 3.0$. Umaima earned a $72$ on the exam. Find the z-score for Umaima's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Umaima's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{72 - {73}}{{3.0}}} $ ${ z \approx -0.33}$ The z-score is $-0.33$. In other words, Umaima's score was $0.33$ standard deviations below the mean.